Write a program to print all the LEADERS in the array. An element is leader if it is greater than all the elements to its right side. And the rightmost element is always a leader. For example int the array {16, 17, 4, 3, 5, 2}, leaders are 17, 5 and 2.
Let the input array be arr[] and size of the array be size.
Method 1 (Simple)
Use two loops. The outer loop runs from 0 to size – 1 and one by one picks all elements from left to right. The inner loop compares the picked element to all the elements to its right side. If the picked element is greater than all the elements to its right side, then the picked element is the leader.
C++
#include<iostream> using namespace std; /*C++ Function to print leaders in an array */void printLeaders(int arr[], int size) { for (int i = 0; i < size; i++) { int j; for (j = i+1; j < size; j++) { if (arr[i] < arr[j]) break; } if (j == size) // the loop didn't break cout << arr[i] << " "; } } /* Driver program to test above function */int main() { int arr[] = {16, 17, 4, 3, 5, 2}; int n = sizeof(arr)/sizeof(arr[0]); printLeaders(arr, n); return 0; } |
Java
class LeadersInArray { /*Java Function to print leaders in an array */ void printLeaders(int arr[], int size) { for (int i = 0; i < size; i++) { int j; for (j = i + 1; j < size; j++) { if (arr[i] < arr[j]) break; } if (j == size) // the loop didn't break System.out.print(arr[i] + " "); } } /* Driver program to test above functions */ public static void main(String[] args) { LeadersInArray lead = new LeadersInArray(); int arr[] = new int[]{16, 17, 4, 3, 5, 2}; int n = arr.length; lead.printLeaders(arr, n); } } |
Python
# Python Function to print leaders in array def printLeaders(arr,size): for i in range(0, size): for j in range(i+1, size): if arr[i]<arr[j]: break if j == size-1: # If loop didn't break print arr[i], # Driver function arr=[16, 17, 4, 3, 5, 2] printLeaders(arr, len(arr)) # This code is contributed by _Devesh Agrawal__ |
C#
// C# program to print // leaders in array using System; class GFG { void printLeaders(int []arr, int size) { for (int i = 0; i < size; i++) { int j; for (j = i + 1; j < size; j++) { if (arr[i] < arr[j]) break; } // the loop didn't break if (j == size) Console.Write(arr[i] + " "); } } // Driver Code public static void Main() { GFG lead = new GFG(); int []arr = new int[]{16, 17, 4, 3, 5, 2}; int n = arr.Length; lead.printLeaders(arr, n); } } // This code is contributed by // Akanksha Rai(Abby_akku) |
PHP
<?php // PHP Function to print // leaders in an array function printLeaders($arr, $size) { for ($i = 0; $i < $size; $i++) { for ($j = $i + 1; $j < $size; $j++) { if ($arr[$i] < $arr[$j]) break; } // the loop didn't break if ($j == $size) echo($arr[$i] . " "); } } // Driver Code $arr = array(16, 17, 4, 3, 5, 2); $n = sizeof($arr); printLeaders($arr, $n); // This code is contributed // by Shivi_Aggarwal ?> |
Output:
17 5 2
Time Complexity: O(n*n)
Method 2 (Scan from right)
Scan all the elements from right to left in an array and keep track of maximum till now. When maximum changes its value, print it.
Below image is a dry run of the above approach:
Below is the implementation of the above approach:
C++
#include <iostream> using namespace std; /* C++ Function to print leaders in an array */void printLeaders(int arr[], int size) { int max_from_right = arr[size-1]; /* Rightmost element is always leader */ cout << max_from_right << " "; for (int i = size-2; i >= 0; i--) { if (max_from_right <= arr[i]) { max_from_right = arr[i]; cout << max_from_right << " "; } } } /* Driver program to test above function*/int main() { int arr[] = {16, 17, 4, 3, 5, 2}; int n = sizeof(arr)/sizeof(arr[0]); printLeaders(arr, n); return 0; } |
Java
class LeadersInArray { /* Java Function to print leaders in an array */ void printLeaders(int arr[], int size) { int max_from_right = arr[size-1]; /* Rightmost element is always leader */ System.out.print(max_from_right + " "); for (int i = size-2; i >= 0; i--) { if (max_from_right <= arr[i]) { max_from_right = arr[i]; System.out.print(max_from_right + " "); } } } /* Driver program to test above functions */ public static void main(String[] args) { LeadersInArray lead = new LeadersInArray(); int arr[] = new int[]{16, 17, 4, 3, 5, 2}; int n = arr.length; lead.printLeaders(arr, n); } } |
Python
# Python function to print leaders in array def printLeaders(arr, size): max_from_right = arr[size-1] print max_from_right, for i in range( size-2, -1, -1): if max_from_right <= arr[i]: print arr[i], max_from_right = arr[i] # Driver function arr = [16, 17, 4, 3, 5, 2] printLeaders(arr, len(arr)) # This code contributed by _Devesh Agrawal__ |
C#
// C# program to find Leaders in an array using System; class LeadersInArray { // C# Function to print leaders // in an array void printLeaders(int []arr, int size) { int max_from_right = arr[size - 1]; // Rightmost element is always leader Console.Write(max_from_right +" "); for (int i = size - 2; i >= 0; i--) { if (max_from_right <= arr[i]) { max_from_right = arr[i]; Console.Write(max_from_right +" "); } } } // Driver Code public static void Main(String[] args) { LeadersInArray lead = new LeadersInArray(); int []arr = new int[]{16, 17, 4, 3, 5, 2}; int n = arr.Length; lead.printLeaders(arr, n); } } // This code is contributed // by Akanksha Rai(Abby_akku) |
PHP
<?php // PHP Function to print // leaders in an array function printLeaders(&$arr, $size) { $max_from_right = $arr[$size - 1]; // Rightmost element // is always leader echo($max_from_right); echo(" "); for ($i = $size - 2; $i >= 0; $i--) { if ($max_from_right <= $arr[$i]) { $max_from_right = $arr[$i]; echo($max_from_right); echo(" "); } } } // Driver Code $arr = array(16, 17, 4, 3, 5, 2); $n = sizeof($arr); printLeaders($arr, $n); // This code is contributed // by Shivi_Aggarwal ?> |
Output:
2 5 17
Time Complexity: O(n)
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